Two wires of the same material have lengths in the ratio `1:2` and their radii are in the ratio `1:sqrt(2)` If they are stretched by applying equal forces, the increase in their lengths will be in the ratio

Options

A. 2
B. `sqrt(2):2`
C. `1:1`
D. `1:2`

Answer

To solve the problem, we will use the formula for the change in length (ΔL) of a wire when a force is applied. The formula is given by:

\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

where:
- \( \Delta L \) is the change in length,
- \( F \) is the applied force,
- \( L \) is the original length of the wire,
- \( A \) is the cross-sectional area of the wire,
- \( Y \) is Young's modulus of the material.

Since both wires are made of the same material, they will have the same Young's modulus \( Y \). We need to find the ratio of the increase in lengths \( \Delta L_1 \) and \( \Delta L_2 \) for the two wires.

### Step 1: Define the lengths and radii of the wires
Let:
- Length of wire 1, \( L_1 = L \)
- Length of wire 2, \( L_2 = 2L \) (since the ratio is 1:2)
- Radius of wire 1, \( R_1 = r \)
- Radius of wire 2, \( R_2 = \sqrt{2}r \) (since the ratio is 1:√2)

### Step 2: Calculate the cross-sectional areas
The cross-sectional area \( A \) of a wire is given by:

\[ A = \pi R^2 \]

Thus, we have:
- Area of wire 1, \( A_1 = \pi R_1^2 = \pi r^2 \)
- Area of wire 2, \( A_2 = \pi R_2^2 = \pi (\sqrt{2}r)^2 = 2\pi r^2 \)

### Step 3: Write the expressions for change in lengths
Using the formula for change in length:

\[ \Delta L_1 = \frac{F \cdot L_1}{A_1 \cdot Y} = \frac{F \cdot L}{\pi r^2 \cdot Y} \]

\[ \Delta L_2 = \frac{F \cdot L_2}{A_2 \cdot Y} = \frac{F \cdot 2L}{2\pi r^2 \cdot Y} = \frac{F \cdot L}{\pi r^2 \cdot Y} \]

### Step 4: Find the ratio of the changes in lengths
Now, we can find the ratio of the changes in lengths:

\[ \frac{\Delta L_1}{\Delta L_2} = \frac{\frac{F \cdot L}{\pi r^2 \cdot Y}}{\frac{F \cdot L}{\pi r^2 \cdot Y}} = \frac{1}{1} = 1 \]

### Conclusion
The increase in their lengths will be in the ratio \( 1:1 \).